# 1567. Maximum Length of Subarray With Positive Product



T: O(n)

S: O(1)

```java
class Solution {
    public int getMaxLen(int[] nums) {
        int n = nums.length;
        int result = 0;
        
        for (int i = 0; i < n; i++) {
            if (nums[i] == 0) {
                continue;
            }
            
            int j = i; // needs cal from i to j
            int negativeCount = 0;
            int k = -1;
            while (j < n && nums[j] != 0) {
                if (nums[j] < 0) {
                    negativeCount++;
                }
                if (negativeCount % 2 == 0) {
                    result = Math.max(result, j - i + 1);
                } else if (k != -1) { // if has first negative index
                    result = Math.max(result, j - k);
                }
                
                if (k == -1 && nums[j] < 0) { // updated first negative one index
                    k = j;
                }
                j++;
            }
            i = j; // move to inner loop's end
        }
        return result;   
    }
}

/*
[0,1,-2,-3,-4]

skip 0

[1 - 2 -3]

[1-2 -3]

[1,-2,-3]

[-3, -4]


another subarray problem but this time how to find the left, right bound?

ex:
     only want this subbarray, so skip 0
   [         ]
[0,1,-2,-3,-4, 0, 1,2, 3]
   i  k     j     [     ]
                  
another thing is to check in this [ ] => 
if the negative count is even: the ans is (j - i + 1)
if the negative count is odd: needs the first negative number's index k, skip this k, so use k+1 to i 
=> ans is (j - (k+1) + 1) = (j-k)


T: O(n)
S: O(1)
*/
```
